## rainbow22 Group Title Find the limit to 0 of.... [(e^2x)-1]/x Is it possible to do this w/out a calculator? one year ago one year ago

1. zzr0ck3r Group Title

do you know l`hopital rule yet?

2. rainbow22 Group Title

no... :(

3. rainbow22 Group Title

You can't have a zero in the denominater. The answer is 2

4. zzr0ck3r Group Title

dont just give answer, help them find answer....I dont know how to solve this wthout lhopitals rule

5. LolWolf Group Title

@zzr0ck3r yeah, I'm trying to find a direct way to do this, without l'Hopital's

6. rainbow22 Group Title

idk, I'm just practicing for a test and this question was online...

7. zzr0ck3r Group Title

you nee to know how to tke a derivative to do this, you will lear that soon

8. rainbow22 Group Title

I know how to take a derivative, but how will that help me solve it?

9. zzr0ck3r Group Title

take the derivative of the num and the the denom. then run the limit

10. zzr0ck3r Group Title

you will have (2e^2x)/1 now run the limi

11. zzr0ck3r Group Title

limit*

12. LolWolf Group Title

Oh, I've got one: Well we know: $e^x-1\sim x\\ x\to 0$ So: $\lim_{x\to 0}\frac{e^{2x}-1}{x}=\lim_{u\to 0}\frac{e^{u}-1}{\frac{1}{2}u}$Where $$u=2x$$ Due to that: $\lim_{x\to 0}\frac{u}{\frac{1}{2}u}=\frac{1}{\frac{1}{2}}=2$